10th CBSE math solution for exercise 7.1 part 2

This page 10th CBSE math solution for exercise 7.1 part 2 is going to provide you solution for every problems that you find in the exercise no 7.1

10th CBSE math solution for exercise 7.1 part 2

(4) Check whether (5,-2) (6,4) and (7,-2) are the vertices of an isosceles triangle.

Solution:

Let the given points as A(5,-2)  B(6,4) and C(7,-2)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Length of the side AB

Here x₁ = 5, y₁ = -2, x₂ = 6  and  y₂ = 4

                 = √(6 - 5)² + (4 - (-2))²

                 = √1² + (4+2)²

                 = √1 + 36

                 = √37

Length of the side BC

Here x₁ = 6, y₁ = 4, x₂ = 7  and  y₂ = -2

                 = √(7 - 6)² + (-2 - 4)²

                 = √1² + (-6)²

                 = √1 + 36

                 = √37

Length of the side CA

Here x₁ = 7, y₁ = -2, x₂ = 5  and  y₂ = -2

                 = √(-5 - 7)² + (-2 - (-2))²

                 = √(-12)² + (-2 + 2)²

                 = √144 + 0

                 = √144

            = 12

AB = BC. Since length of two sides are equal. So it forms an isosceles triangle.


In the page 10th CBSE math solution for exercise 7.1 part 2 we are going to see the solution of next problem

(5) In a classroom 4 friends are seated at the points A,B,C and D as shown in fig 7.8.Champa and Chameli walk into the class and after observing for a few minutes champa asks Chameli,"Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.

Solution:

It can be observed that A (3,4) , B (6,7), C(9,4) and D(6,1) are the position of these 4 friends.

Length of AB

Here x₁ = 3, y₁ = 4, x₂ = 6  and  y₂ = 7

                 = √(6 - 3)² + (7 - 4)²

                 = √(3)² + (3)²

                 = √9 + 9

                 = √18

            = 3√2

Length of BC

Here x₁ = 6, y₁ = 7, x₂ = 9  and  y₂ = 4

                 = √(9 - 6)² + (4 - 7)²

                 = √(3)² + (-3)²

                 = √9 + 9

                 = √18

            = 3√2

Length of CD

Here x₁ = 9, y₁ = 4, x₂ = 6  and  y₂ = 1

                 = √(6 - 9)² + (1 - 4)²

                 = √(-3)² + (-3)²

                 = √9 + 9

                 = √18

            = 3√2

Length of DA

Here x₁ = 6, y₁ = 1, x₂ = 3  and  y₂ = 4

                 = √(3 - 6)² + (4 - 1)²

                 = √(-3)² + (3)²

                 = √9 + 9

                 = √18

            = 3√2

Length of diagonal AC

Here x₁ = 3, y₁ = 4, x₂ = 9  and  y₂ = 4

                 = √(9 - 3)² + (4 - 4)²

                 = √6² + 0²

                 = √36

                 = 6

Length of diagonal BD

Here x₁ = 6, y₁ = 7, x₂ = 6  and  y₂ = 1

                 = √(6 - 6)² + (1 - 7)²

                 = √0² + (-6)²

                 = √36

                 = 6

It can be observed that all sides of this quadrilateral ABCD are the same length and also the diagonals are of the same length.

Therefore,ABCD is a square and hence,Champa was correct.
10th CBSE math solution for exercise 7.1 part 2




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